chapter three: infrared spectroscopy (ir)

Dr.Abdulhadi Kadhim.

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Spectroscopy

Ch.3

الجامعة التكنولوجیة

الدكتور عبد الھادي كاظم -:قسم العلوم التطبیقیة استاذ المادة

فرع الفیزیاء التطبیقیة

المرحلة الرابعة

Chapter three: Infrared Spectroscopy

(IR)


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Spectroscopy

Ch.3

Infrared Spectroscopy (IR)

IR

Vibrational energy of diatomic molecule:- We are all familiar with the vertical oscillations of amass (m) connected to a stretched spring of a force constant k whose ether end is fixed.

The simple harmonic motion with a fundamental frequency:-

° =

If two masses in a diatomic molecule m1 and m2 we used the reduced mass

∴ = in quantum mechanically, the vibrational energy is given by

° = + ° υ = 0,1,2,3−−−−

Where υ is the vibrational quantum No.

The energy in Cm-1

= = ( + ) ° = ( + ) °\

Where ° the freq. in cm-1. These energy levels are equally spaced

and the energy of lowest state ° = °\ ∶

Fundamental region (2.5-15.4µm) Near IR (0.75-2.5µm)

Far IR (15.4-microwave)


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An harmonic oscillator In addition to rotational motion , molecule has vibrational motion which is an harmonic. Hence a further correction to the centrifugal distortion to be made to account for the vibrational excitation of the molecule.

The empirical equation of the potential energy of the diatomic molecule is given by V = D 1− e ( ) This equation called morse equation. V = morse potential. This potential represent the actual curve to a very good degree of approximation except when r=0 → Vm gives a high finite value while the actual value is infinity.

v At r→ ∞ → Vm→De De: dissociation energy. v At r=re → Vm→0

V

D

r re


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Let x=r-re x<< 1 very small = 1− + 2! − 3! + ⋯… … … … … … .. Or = 1− + ( ) ! − ( ) ! + ⋯… … … … … … .. We take the first two term = 1 − = 1 − ( ) = [1 − ( − )] = [− ( − )] = ( − ) ≅ 1 −

≅ ( − ) = (2 )( − ) = \( − )

Where \: 2a2De=force constant for small displacement. = \ = in (Hz)

: freq. for small displacement.

Substitution Vm solving this eq. using pert buation theory we

gets G(v) = ω v + − ω x v +

ω x :an harmonicity correction and constant.

ω >> ω x

The effects of an harmonicity:-

1- The energy levels are not equally spaced. 2- Selection rules for allowed transition

∆v=±1,±2,±3,…… v When ∆v=±1→ G(v1)= G(1)- G(0)


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First fundamental freq.= ω -2ω x v When ∆v=±2 → G(v2)= G(2)- G(0) First over ton fundamental freq.=2 ω -6ω x D = ω 4ω x

We can derive the equation of D G(v) = ω v + 12 − ω x 12 − − −−(1)

The maximum energy at = 0 Then G(v) =De ( ) = ω -2ω x v + = ω − 2vω x − ω x

2vω x =ω − ω x ∴ = ω ω ω

Sub in eq. (1)

∴ ( ) = + − +

= − + 2 − − + 2 2

= − ( )


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= − = D = Dissociation energy

Example(1):-

The fundamental vibration spectrum of Co molecule at the wave number 2138 Cm-1 and the line which represent by the transition v=2→ v=3 at wave number 2091 Cm-1 calculate

1. The harmonic constant ( ). 2. The unharmonic constant ( ). 3. The Dissociation energy.

The solution:-

G(v) = ω v + 12 − ω x 12

G(v1)= G(1)- G(0)

( ) = 1 + − 1 + − 0 + − 0 + ( ) = 32 − 94 − 12 + 14 ( ) = − 2 2138 = − 2 − − −−(1)

By the same way

G(v2)= G(3)- G(2)


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Ch.3 G(v ) = 72ω − 494 ω x − 52ω − 252 ω x = ω − 6ω x ∴ 2091 = − 6 − − − −(2)

∴ω x = cm = 11.75 cm

Sub in eq.(1)

2138 cm-1 = (2 × 11.75 )

∴ = 2161.5 = 4 = (2161.5) 4 × 11.75 = 4.44 × 10 9.94 × 10

= 9.94 × 10 1 × 3 × 10 = 29.82 × 10 1

D = 29.82 × 10 1sec × 6.63 × 10 J. S= 147.7 × 10 Joul ∴ D = 147.7 × 10 1.6 × 10 = 123.56 × 10 ev = 12.356 ev

2091 = − 6 −− − −(2) 2138 = − 2 −− − −(1)

-47 cm-1 = -4

بالطرح


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Ch.3

H.W.(1):-

The absorption spectrum of HF molecule in the IR region , the first line at 3961 cm-1 and the second line at 7751 cm-1 calculate

1. The dissociation constant 2f. 2. The Dissociation energy D .

Population of vibration energy levels According to Boltzman distribution

= ∆ / = /

∆E For first three transitions

(1) V=0 → V=1 ( Fundamental frequency ) ∆V=+1 ∆E = εv=1 - εv=0

∆E = 1 + − 1 + − − ∆E= (1− 2 )

(2) V=0 → V=2 ∆V=+2

∆E=2 − 6

∆E=2 (1 − 3 ) (3) V=0 → V=3 ∆V=+3

∆E=3 − 12

∆E=3 (1 − 3 )


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Example(2):-

HCl molecule is absorb the radiation at 2885.9 cm-1 by using Boltzman distribution calculate the relative Number for first vibrational level and the ground level at 25oC suppose the No. of molecule at ground level is one.

The solution:-

= ∆ / = /

= . × . × × × / . × × = 9.245 × 10

H.W.(2):-

HCl molecule is absorb the IR radiation at fundamental vibrational frequency =2890 cm-1 calculate the force constant ?

H.W(3):- Calculate the force constant for OH which can be the spectrum absorb at IR region .

Infrared selection Rules

The selection Rule of IR if the vibration quantum Number change

1. ∆V=±1 under harmonic approximation. 2. ∆V=±1, ±2, ±3 for anharmonic oscillator.


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Over tones frequency:-

On other transitions may be take place furthermore the fundamental frequency, at 2V1, 3V1, 4V1

1. 2V1, 3V1, 4V1 for transition began from v=0. 2. 2V2, 3V2, 4V2 for transition began from v=1. 3. 2V3, 3V3, 4V3 for transition began from v=2.

And the frequency of the 1st overtone = 2 e(1-3xe) The absorption due to these transition called (( Over tones frequencies))

The following figure represent the overtones frequencies for Hcl molecule in IR region.

Zero point energy = − 1 −

V (cm-

1) 0

V =0 →1 Fundamental frequencies

5000

First over ton freq.

V =0 →2

2nd over ton freq.

V =0 →3

10000

3rd over ton freq.

V =0 →4

4th over ton freq.

V =0 →5


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Combination bands and differences bands:-

The selection rules allowed by other transitions to produce combination bands and difference bands. The combination bands

is arise by added two or more fundamentals frequencies or over tons frequencies such as:- + , 2 + , + +

These frequencies become allowed but the intensity is very low.

The difference bands is similar to combination bands but the transition appear in complex spectra. − , 2 − , + −

Born-Oppenheimer principle ((The electrons movements is high speed compared with nuclei movements , then can be considered the nuclei is constant when studies the electronic properties of molecules and the energy of electrons is independent on nuclei energy and the total energy is equal to sum of vibration energy))

Plus Rotational energy plus electronic energy . = . + . + . Vibration – Rotation spectroscopy of diatomic molecules The vibration and rotation motion happen in any molecule at the same time, and for any vibration level consist from


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many Rotational levels. The following figure explain the first two vibrational level v=0 and v=1 and the rotational levels.

The vibration – rotation spectrum produce by transitions from rotational levels of vibration level like v=0 to another rotational levels of vibration level like v=1 and the selection rule for vibration- Rotation spectra

∆V=±1 , ∆J =±1

The energy of vibration-Rotation for diatomic molecule

E(J,V) = E(J) + E(V)

= BJ(J+1) + + ° = ℎ ( + 1) + + + +

v=0 , v=1لدینا حالة الانتقال بین مستویین مثلا

v=1

4 2B 2B

vo R-branch P-branch

0 1

2

3

4 /J

“J

0 1

2

3

4

∆ J=+1 ∆ J=-1

v=0


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= \ \ + − \\ \\ + + \ + ± \\ +

The vibrational-rotational absorption spectrum

Consists of two groups of equally spaced lines(R+P)branches with a gap between them, the center of this gap is υo But experimentally it is found that the spacing is not constant , this is due to the vibration- rotation coupling.

When take the distortion effect ( centrifugal distortion ) in account the = , = ( + 1) − ( + 1) + + − +

Consider the vibrational rotational transition υ=0→υ=1

Assuming B and D the same for both υ=0, and υ=1 states and denoting upper state by single prime and lower state by double prime = \ \ + 1 − \\ \\ + 1 − \ \ + 1 − \\ \\ + 1 + 32 − 94 − 12 − 14

= (1 − 2 ) + \ − \\ \ + \\ + 1 − \ \ + 1 − \\ \\ + 1 -(1)

Use of the selection rule ∆J=+1 ∴ J\-J\\=1 gives = + 2 \\ + 1 − 4 \\ + 1 , = 0,1,2,− −− −−

∆J=-1 ∴ J\-J\\=-1 gives = − 2 \ + 1 + 4 \ + 1 , = 0,1,2,−− −− −

Where = (1 − 2 ) which is the freq. of the υ=0→υ=1transition


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Spectroscopy

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Lines corresponding to ∆J=-1 called P branch.

Lines corresponding to ∆J=+1 called R branch. , = + − (m=J\\+1 and J\+1)

usually D is extremely small and in such cases

, = +

-:(3)Example

The normal modes of vibration of Co2 molecule are υ1=1330 cm-1, υ2=667 cm-1, υ3=2349 cm-1 and υ4=2349 cm-1 evaluate the zero point energy of Co2 molecule.

-Solution:

The Co2 molecule has 4 normal modes of vibration the symmetric bending mode υ2 is double degenerate.

The zero point energy Eo of Co2= ℎ ∑ \ ∑ \ = (1330 + 667 + 667 + 2349) = 5013 × 100

∴ ° = 6.6 × 10 × 3 × 10 × 5013 × 100 = 4.98 × 10 = 0.311


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-: H.W.(4)

The equilibrium vibration frequency of the Iodine molecule is υs cm-1 and the anharmonicity constant = 0.003

What is the intensity of hot band υ=1→υ=2 relative to that of the fundamental υ=0→υ=1 , if the temperature is 300 oK ?

Note: - The hot band occurs in the temperature higher than the room temp.

-:(4)Example

The fundamental and first over tone transition of 14N16O are centered at 1876.06 cm-1 and 3724.20 cm-1 respectively. Evaluate the equilibrium vibration frequency, the anharmonicity constant, zero point energy and force constant of molecule?

-Solution: = (1− 2 ) = 1876.06 cm over tone = 2 (1− 3 ) = 3724.20 cm = 7.332 × 10 \ = 1903.98

Zero point energy ° = \ 1−


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Ch.3 = × 1903.98 (1 − 0.003666) = 948.5

Mass of 14N =23.25x10-27Kg

Mass of 16O =26.56x10-27Kg = . × . . . × 10 = 12.3975 × 10 Kg

= 4 \ = 4 × 12.3975 × 10 × 9 × 10 × (904 × 100) = 1597 N/m

H.W.(5):- Estimate the position of the band center and B value of the HCl from the few lines of the P and R branches listed below:-

line line 2906.24 Ro 2865.1 P1 2925.90 R1 2843.62 P2 2944.90 R2 2821.56 P3 2963.29 R3 2798.94 P4


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Spectroscopy

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-:(5)Example

The fundamental band for HCl is centered at 2886 cm-1 . Assuming that the inter nuclear distance is 1.276 oA. Calculate the wave No. of the first two lines of each of the P and R branches of HCl.

-Solution:

The reduced mass of HCl = ( . )×( . )( . ) ( . ) = 1.627 × 10 = = . × × . × ×( . × ) × = 10.6129 , = + 2 = ∓1,∓2,∓3−−−−− = 2886− 2 × 10.6129 = 2864.77 = 2886− 4 × 10.6129 = 2843.55 = 2886 + 2 × 10.6129 = 2907.23 = 2886 + 4 × 10.6129 = 2928.45


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-IR Spectrometer:

Consist from the following parts:-

(1) Source of radiation:-

Rod from silicon carbide heated to 1500 oC.

(2) Two mirrors to produce two beams, the first beam reference beam and the second beam is sample beam.

(3) Cell of sample: - made from nacl.

(4) Grating or lit row mount prism.

(5)The sample may be solid or liquid ((If the sample is gas must be used special cells of some millimeters thickness))

-IR applications:

(1) Diagnosis the structure of organic compounds.

(2) Determine the purity of compounds.

(3) Quantitative analysis for different compounds.

(4) A study of chemical interactions.

(5) A study of Hydrogen bonding.

(6) Analysis of orientation and geometric shapes for small molecules in gas state.

(7) A study of polymers chemical.


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Orational Infrared spectrum for polyatomic molecules

1- Linear Polyatomic molecules When molecule contain N of atoms then the axis (x, y, z) this leads to 3N of axis this molecule need or called 3N freedom degree.

This means

(3N) to characterized the position or motion of center of molecule mass, and (2N) for rotational motion then:-

3N-5 No. of degree of freedom for vibrational motion

N:-No. of atoms

For Diatomic molecules 3N-5=1 means the No. of vibrational motion

equal to one.

For Linear molecule have three atoms such as Co2 3N-5=3x3-5=4 No. of vibrational ( Fundamental vibration) two as parallel of axis bond and two represented the atom motion outside of axis bond .

The first one called ( stretching ) and the second called (bending)

Starching Starching


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2- Non Linear Polyatomic molecules:- The No. of vibration equal to 3N-6 Three for transition motion and Three for rotational motion.

-Example:

Water molecule H2o

The No. of vibration 3N-6=3

The energy of stretching is higher than the energy of bending vibration.

The rotational energy level occurs as a result of rotation of molecules around axis perpendicular on has symmetry axis and the axis of symmetry has been neglected due to the mass of atom insure in nuclear and radius of nuclear is very small compared with atomic radius this leads to the moment of inertia due to electron rotation around the axis.

O

H H

O

H H

O

H H


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The Features of Rotational levels:-

(1) The separated energy between the spectrum lines about 10-3ev.

(2) The wave length about ( 0.1mm-several cm).

(3) Lie in micro wave region.

(4) The moment of inertia equal to I=µr2 .

(5) The energy = Iω = ℏ J(J + 1).

(6) Appear clearly in permanent dipole molecule and does not appear in symmetrical molecule like H2,N2 .

(7) The selection rules ∆J=±1 and µ≠0 .

(8) The energy in room temp. greater than rotational energy

( KT= 2.6 x 10-2 ev) .

(9) Most rotational spectrum shown in absorption spectrum and appear in cases and vapor.

chapter three: infrared spectroscopy (ir)

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